/*
 *                               POK header
 *
 * The following file is a part of the POK project. Any modification should
 * be made according to the POK licence. You CANNOT use this file or a part
 * of a file for your own project.
 *
 * For more information on the POK licence, please see our LICENCE FILE
 *
 * Please follow the coding guidelines described in doc/CODING_GUIDELINES
 *
 *                                      Copyright (c) 2007-2021 POK team
 */

/* @(#)e_jn.c 5.1 93/09/24 */
/*
 * ====================================================
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
 *
 * Developed at SunPro, a Sun Microsystems, Inc. business.
 * Permission to use, copy, modify, and distribute this
 * software is freely granted, provided that this notice
 * is preserved.
 * ====================================================
 */

/*
 * __ieee754_jn(n, x), __ieee754_yn(n, x)
 * floating point Bessel's function of the 1st and 2nd kind
 * of order n
 *
 * Special cases:
 *	y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
 *	y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
 * Note 2. About jn(n,x), yn(n,x)
 *	For n=0, j0(x) is called,
 *	for n=1, j1(x) is called,
 *	for n<x, forward recursion us used starting
 *	from values of j0(x) and j1(x).
 *	for n>x, a continued fraction approximation to
 *	j(n,x)/j(n-1,x) is evaluated and then backward
 *	recursion is used starting from a supposed value
 *	for j(n,x). The resulting value of j(0,x) is
 *	compared with the actual value to correct the
 *	supposed value of j(n,x).
 *
 *	yn(n,x) is similar in all respects, except
 *	that forward recursion is used for all
 *	values of n>1.
 *
 */

#ifdef POK_NEEDS_LIBMATH

#include "math_private.h"
#include "namespace.h"
#include <libm.h>

static const double invsqrtpi =
                        5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
    two = 2.00000000000000000000e+00,               /* 0x40000000, 0x00000000 */
    one = 1.00000000000000000000e+00;               /* 0x3FF00000, 0x00000000 */

static const double zero = 0.00000000000000000000e+00;

double __ieee754_jn(int n, double x) {
  int32_t i, hx, ix, lx, sgn;
  double a, b, temp, di;
  double z, w;

  temp = 0;
  /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
   * Thus, J(-n,x) = J(n,-x)
   */
  EXTRACT_WORDS(hx, lx, x);
  ix = 0x7fffffff & hx;
  /* if J(n,NaN) is NaN */
  if ((ix | ((uint32_t)(lx | -lx)) >> 31) > 0x7ff00000)
    return x + x;
  if (n < 0) {
    n = -n;
    x = -x;
    hx ^= 0x80000000;
  }
  if (n == 0)
    return (__ieee754_j0(x));
  if (n == 1)
    return (__ieee754_j1(x));
  sgn = (n & 1) & (hx >> 31); /* even n -- 0, odd n -- sign(x) */
  x = fabs(x);
  if ((ix | lx) == 0 || ix >= 0x7ff00000) /* if x is 0 or inf */
    b = zero;
  else if ((double)n <= x) {
    /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
    if (ix >= 0x52D00000) { /* x > 2**302 */
      /* (x >> n**2)
       *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
       *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
       *	    Let s=sin(x), c=cos(x),
       *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
       *
       *		   n	sin(xn)*sqt2	cos(xn)*sqt2
       *		----------------------------------
       *		   0	 s-c		 c+s
       *		   1	-s-c 		-c+s
       *		   2	-s+c		-c-s
       *		   3	 s+c		 c-s
       */
      switch (n & 3) {
      case 0:
        temp = cos(x) + sin(x);
        break;
      case 1:
        temp = -cos(x) + sin(x);
        break;
      case 2:
        temp = -cos(x) - sin(x);
        break;
      case 3:
        temp = cos(x) - sin(x);
        break;
      }
      b = invsqrtpi * temp / sqrt(x);
    } else {
      a = __ieee754_j0(x);
      b = __ieee754_j1(x);
      for (i = 1; i < n; i++) {
        temp = b;
        b = b * ((double)(i + i) / x) - a; /* avoid underflow */
        a = temp;
      }
    }
  } else {
    if (ix < 0x3e100000) { /* x < 2**-29 */
      /* x is tiny, return the first Taylor expansion of J(n,x)
       * J(n,x) = 1/n!*(x/2)^n  - ...
       */
      if (n > 33) /* underflow */
        b = zero;
      else {
        temp = x * 0.5;
        b = temp;
        for (a = one, i = 2; i <= n; i++) {
          a *= (double)i; /* a = n! */
          b *= temp;      /* b = (x/2)^n */
        }
        b = b / a;
      }
    } else {
      /* use backward recurrence */
      /* 			x      x^2      x^2
       *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
       *			2n  - 2(n+1) - 2(n+2)
       *
       * 			1      1        1
       *  (for large x)   =  ----  ------   ------   .....
       *			2n   2(n+1)   2(n+2)
       *			-- - ------ - ------ -
       *			 x     x         x
       *
       * Let w = 2n/x and h=2/x, then the above quotient
       * is equal to the continued fraction:
       *		    1
       *	= -----------------------
       *		       1
       *	   w - -----------------
       *			  1
       * 	        w+h - ---------
       *		       w+2h - ...
       *
       * To determine how many terms needed, let
       * Q(0) = w, Q(1) = w(w+h) - 1,
       * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
       * When Q(k) > 1e4	good for single
       * When Q(k) > 1e9	good for double
       * When Q(k) > 1e17	good for quadruple
       */
      /* determine k */
      double t, v;
      double q0, q1, h, tmp;
      int32_t k, m;
      w = (n + n) / (double)x;
      h = 2.0 / (double)x;
      q0 = w;
      z = w + h;
      q1 = w * z - 1.0;
      k = 1;
      while (q1 < 1.0e9) {
        k += 1;
        z += h;
        tmp = z * q1 - q0;
        q0 = q1;
        q1 = tmp;
      }
      m = n + n;
      for (t = zero, i = 2 * (n + k); i >= m; i -= 2)
        t = one / (i / x - t);
      a = t;
      b = one;
      /*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
       *  Hence, if n*(log(2n/x)) > ...
       *  single 8.8722839355e+01
       *  double 7.09782712893383973096e+02
       *  long double 1.1356523406294143949491931077970765006170e+04
       *  then recurrent value may overflow and the result is
       *  likely underflow to zero
       */
      tmp = n;
      v = two / x;
      tmp = tmp * __ieee754_log(fabs(v * tmp));
      if (tmp < 7.09782712893383973096e+02) {
        for (i = n - 1, di = (double)(i + i); i > 0; i--) {
          temp = b;
          b *= di;
          b = b / x - a;
          a = temp;
          di -= two;
        }
      } else {
        for (i = n - 1, di = (double)(i + i); i > 0; i--) {
          temp = b;
          b *= di;
          b = b / x - a;
          a = temp;
          di -= two;
          /* scale b to avoid spurious overflow */
          if (b > 1e100) {
            a /= b;
            t /= b;
            b = one;
          }
        }
      }
      b = (t * __ieee754_j0(x) / b);
    }
  }
  if (sgn == 1)
    return -b;
  else
    return b;
}

double __ieee754_yn(int n, double x) {
  int32_t i, hx, ix, lx;
  int32_t sign;
  double a, b, temp;

  temp = 0;
  EXTRACT_WORDS(hx, lx, x);
  ix = 0x7fffffff & hx;
  /* if Y(n,NaN) is NaN */
  if ((ix | ((uint32_t)(lx | -lx)) >> 31) > 0x7ff00000)
    return x + x;
  if ((ix | lx) == 0)
    return -one / zero;
  if (hx < 0)
    return zero / zero;
  sign = 1;
  if (n < 0) {
    n = -n;
    sign = 1 - ((n & 1) << 1);
  }
  if (n == 0)
    return (__ieee754_y0(x));
  if (n == 1)
    return (sign * __ieee754_y1(x));
  if (ix == 0x7ff00000)
    return zero;
  if (ix >= 0x52D00000) { /* x > 2**302 */
    /* (x >> n**2)
     *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
     *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
     *	    Let s=sin(x), c=cos(x),
     *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
     *
     *		   n	sin(xn)*sqt2	cos(xn)*sqt2
     *		----------------------------------
     *		   0	 s-c		 c+s
     *		   1	-s-c 		-c+s
     *		   2	-s+c		-c-s
     *		   3	 s+c		 c-s
     */
    switch (n & 3) {
    case 0:
      temp = sin(x) - cos(x);
      break;
    case 1:
      temp = -sin(x) - cos(x);
      break;
    case 2:
      temp = -sin(x) + cos(x);
      break;
    case 3:
      temp = sin(x) + cos(x);
      break;
    }
    b = invsqrtpi * temp / sqrt(x);
  } else {
    uint32_t high;
    a = __ieee754_y0(x);
    b = __ieee754_y1(x);
    /* quit if b is -inf */
    GET_HIGH_WORD(high, b);
    for (i = 1; i < n && high != 0xfff00000; i++) {
      temp = b;
      b = ((double)(i + i) / x) * b - a;
      GET_HIGH_WORD(high, b);
      a = temp;
    }
  }
  if (sign > 0)
    return b;
  else
    return -b;
}
#endif
